20.3. Force on a moving charge
A subsection of Physics, 9702, through 20. Magnetic fields
Listing 10 of 105 questions
An electron having charge –q and mass m is accelerated from rest in a vacuum through a potential difference V. The electron then enters a region of uniform magnetic field of magnetic flux density B, as shown in . path of electron uniform magnetic field into plane of paper The direction of the uniform magnetic field is into the plane of the paper. The velocity of the electron as it enters the magnetic field is normal to the magnetic field. The radius of the circular path of the electron in the magnetic field is r. Explain why the path of the electron in the magnetic field is the arc of a circle. Show that the magnitude p of the momentum of the electron as it enters the magnetic field is given by p = (2mqV ). The potential difference V is 120 V. The radius r of the circular arc is 7.4 cm. Determine the magnitude B of the magnetic flux density. B = T The potential difference V in is increased. The magnetic flux density B remains unchanged. By reference to the momentum of the electron, explain the effect of this increase on the radius r of the path of the electron in the magnetic field.
9702_s17_qp_41
THEORY
2017
Paper 4, Variant 1
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An electron having charge –q and mass m is accelerated from rest in a vacuum through a potential difference V. The electron then enters a region of uniform magnetic field of magnetic flux density B, as shown in . path of electron uniform magnetic field into plane of paper The direction of the uniform magnetic field is into the plane of the paper. The velocity of the electron as it enters the magnetic field is normal to the magnetic field. The radius of the circular path of the electron in the magnetic field is r. Explain why the path of the electron in the magnetic field is the arc of a circle. Show that the magnitude p of the momentum of the electron as it enters the magnetic field is given by p = (2mqV ). The potential difference V is 120 V. The radius r of the circular arc is 7.4 cm. Determine the magnitude B of the magnetic flux density. B = T The potential difference V in is increased. The magnetic flux density B remains unchanged. By reference to the momentum of the electron, explain the effect of this increase on the radius r of the path of the electron in the magnetic field.
9702_s17_qp_43
THEORY
2017
Paper 4, Variant 3
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Explain how a uniform magnetic field and a uniform electric field may be used as a velocity selector for charged particles. Particles having mass m and charge +1.6 × 10–19 C pass through a velocity selector. They then enter a region of uniform magnetic field of magnetic flux density 94 mT with speed 3.4 × 104 m s–1, as shown in . path of charged particle velocity selector 15.0 cm uniform magnetic field into page The direction of the uniform magnetic field is into the page and normal to the direction in which the particles are moving. The particles are moving in a vacuum in a circular arc of diameter 15.0 cm. Show that the mass of one of the particles is 20 u. On , sketch the path in the uniform magnetic field of a particle of mass 22 u having the same charge and speed as the particle in .
9702_s18_qp_42
THEORY
2018
Paper 4, Variant 2
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A spherical oil drop has a radius of 1.2 × 10−6 m. The density of the oil is 940 kg m−3. Show that the mass of the oil drop is 6.8 × 10−15 kg. The oil drop is charged. Explain why it is impossible for the magnitude of the charge to be 8.0 × 10−20 C. The charged oil drop in is in a vacuum between two horizontal metal plates, as illustrated in . 8.0 mm uniform electric field, field strength 2.1 × 105 V m–1 metal plate metal plate oil drop, mass 6.8 × 10–15 kg + V The plates are separated by a distance of 8.0 mm. The electric field between the plates is uniform and has a field strength of 2.1 × 105 V m−1. The oil drop moves vertically downwards with a constant speed. Calculate the potential difference V between the plates. V = V Explain how the motion of the oil drop shows that it is in equilibrium. Determine the charge on the oil drop. charge = C sign of charge The magnitude of the potential difference between the plates in is decreased. Explain why the oil drop accelerates downwards. Describe the change to the pattern of the field lines (lines of force) representing the uniform electric field as the potential difference decreases. Two types of force, X and Y, can act on an oil drop when it is in air, but cannot act on an oil drop when it is in a vacuum. Force X can act on an oil drop when it is stationary or when it is moving. Force Y can only act on an oil drop when it is moving. State the name of: force X force Y.
9702_s19_qp_21
THEORY
2019
Paper 2, Variant 1
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An electron is travelling in a vacuum at a speed of 3.4 × 107 m s–1. The electron enters a region of uniform magnetic field of flux density 3.2 mT, as illustrated in . 30° region of uniform magnetic flux density 3.2 mT electron speed 3.4 × 107 m s–1 The initial direction of the electron is at an angle of 30° to the direction of the magnetic field. When the electron enters the magnetic field, the component of its velocity vN normal to the direction of the magnetic field causes the electron to begin to follow a circular path. Calculate: vN vN = m s–1 the radius of this circular path. radius = m State the magnitude of the force, if any, on the electron in the magnetic field due to the component of its velocity along the direction of the field. Use information from and to describe the resultant path of the electron in the magnetic field.
9702_s19_qp_42
THEORY
2019
Paper 4, Variant 2
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An electron is travelling at speed v in a straight line in a vacuum. It enters a uniform magnetic field of flux density 8.0 × 10–4 T. Initially, the electron is travelling at right angles to the magnetic field, as illustrated in . path of electron region of uniform magnetic field The path of the electron in the magnetic field is an arc of a circle of radius 6.4 cm. State and explain the direction of the magnetic field. Show that the speed v of the electron is 9.0 × 106 m s–1. A uniform electric field is now applied in the same region as the magnetic field. The electron passes undeviated through the region of the two fields, as illustrated in . path of electron region of uniform electric and magnetic fields On , mark with an arrow the direction of the uniform electric field. Use data from to calculate the magnitude of the electric field strength. field strength = N C–1 The electron in is now replaced by an α‑particle travelling at the same speed v along the same initial path as the electron. Describe and explain the shape of the path in the region of the magnetic and electric fields.
9702_s20_qp_42
THEORY
2020
Paper 4, Variant 2
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Questions Discovered
105